1
$\begingroup$

If $A \cup B = A$ then $A$ is a subset of $A$ and $B$ is a subset of $A$. Thus $A \cup B = A$.

If $B$ is a subset of $A$ then it follows that $A \cup B$ is a subset of $A$.

My solution. It seems very straightforward and simple. But is it valid?

$\endgroup$
1
  • 1
    $\begingroup$ Proper subset of $A$ usually means a subset of $A$ which is not $A$ itself, though the statement is obviously true for $B=A$ as well. $\endgroup$
    – Berci
    Commented May 5, 2014 at 21:50

3 Answers 3

2
$\begingroup$

Its a good start! The key to a good proof is a sequence of if..then, therefore...statements that lead from the premise to the conclusion. And remember to use the definitions.

Also keep in mind that we have to prove both of these statements:

  1. if $A \cup B = A$ then $B \subseteq A$
  2. if $B \subseteq A$ then $A \cup B = A$

For the first one, try something like:

If $A \cup B = A$ then by the definition of union, $ A = \{x : x \in A\text{ or } x \in B\}$. So for each $x \in B$, $x \in A$. In other words, $B \subseteq A$.

Can come up with something similar for the second one?

$\endgroup$
2
  • $\begingroup$ Thanks for the advice! Here goes: If B is a subset of A then A = {x : x an element of A or x an element of B}. Then for each x that is an element of B, that same x is an element of A. Therefore, A U B = A as A is a subset of A U B and A U B is a subset of A. $\endgroup$
    – user122661
    Commented May 5, 2014 at 22:18
  • $\begingroup$ Thats great! I might switch sentence 1 and 2 and drop the last sentence to get ... "If B is a subset of A then each x that is an element of B, that same x is an element of A. So A = {x : x an element of A or x an element of B. In other words, $A = A \cup B$." $\endgroup$
    – joebloggs
    Commented May 5, 2014 at 22:25
1
$\begingroup$

First assume $A \cup B=A$

That means $x \in A \cup B \iff x \in A$ so: $$x\in B \implies x \in A \cup B \implies x \in A$$ The later statement is equivalent to $A \subset B$

Second assume $B \subset A$

That means $x \in B \implies x \in A$.

First what is obvious:

$$x \in A \implies x \in A \cup B$$

The later statement means $A \subset A \cup B$

Second for proof $A \cup B \subset A $

First if $x \in A \cup B$ then $x \in A$ or $x \in B$:

$x \in A \implies x \in A$

$x \in B \implies x \in A$ (by hipothesis)

So:

$x \in A \cup B \implies x \in A \vee x \in B \implies x\in A$

$\endgroup$
0
$\begingroup$

If $A \cup B = A$, then there is no element $e \in B$ such that $e \notin A$. Thus $B$ contains only elements that are also in $A$ (if at all: $B$ may be empty), which means that $B \subseteq A$.

If $B \subseteq A$; $A \cup B $ contains all elements of $A$ and $B$. But since $B \subseteq A$, there is no element $e \in B$ such that $e \notin A$, which means that $A \cup B = A$ since $B$ does not add any new elements to this set.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .